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In the reversible reaction A + B `hArr` C + D, the concentration of each C and D at equilobrium was `0.8` mole/litre, then the equilibrium constant `K_(c)` will beA. `6.4`B. `0.64`C. `1.6`D. `16.0` |
Answer» Correct Answer - D Suppose 1 mole of A and B is each taken, then `0.8` mole/litre of C and D each formed remaining concentration of A and B will be `(1 - 0.8) = 0.2` mole/litre. `K_(c) = ([C][D])/([A][B]) = (0.8xx0.8)/(0.2xx0.2) = 16` |
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